An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge −Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. Show that the electric field inside this atom is : Ein=Ze4πϵ0(1r^2−rR^3). b. What is the electric field at the surface of the atom? Is this the expected value? Explain.c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field at r = R/2?

Answers

Answer 1
Answer:

Answer:

Part a)

E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))

Part b)

E = 0

Yes it is the expected value of electric field at the surface of an atom

Part c)

E = 4.64 * 10^(13) N/C

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

E = (kq)/(r^2)

E_1 = (Ze)/(4\pi \epsilon_0 r^2)

now charge due to electrons inside a radius "r" is given as

q = (-Ze r^3)/(R^3)

now we will have electric field given as

E_2 = ((-Zer^3)/(R^3))}{4\pi\epsilon_0 r^2}

now net electric field is given as

E = E_1 + E_2

E = (Ze)/(4\pi \epsilon_0 r^2) - (Zer)/(4\pi \epsilon_0 R^3)

E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))

Part b)

At the surface of an atom

r = R

E = 0

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

r = (R)/(2)

so we will have

E = 92(1.6 * 10^(-19)) * (9 * 10^9)((4)/(R^2) - (1)/(2R^2))

E = (4.64 * 10^(-7))/((0.10 * 10^(-9))^2)

E = 4.64 * 10^(13) N/C


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A flat (unbanked) curve on a highway that has a radius of 50 m. A car rounds the curve. The car has mass 4,907 kg. The static coefficient of friction between the curve and the car is 0.35. What is the maximum speed of the car to prevent sliding?

Answers

Answer:

maximum speed of the car to prevent sliding is 13.1m/s

Explanation:

Given data

Radius of curve r=50m

Mass of car m=4907kg

Coefficient of friction u=0.35

Limiting for R=?

Hence limiting force R=ma

R=4907*9.81

R=48137.7N

We know that the force to overcome friction is

F=uR

Hence

F=0.35*48137.7

F=16848.2N

Centripetal force along the curve is given as

Fc=mv²/r

Fc = centripetal force

m = mass

v = velocity

r = radius

To solve for velocity we have to equate both force required to overcome friction and the centripetal force

Fc=mv²/r=F=uR

mv²/r=uR

Making velocity subject of formula we have

v²=u*r*R/m

v²=(0.35*50*48137.7)/4907

v²=842409.75/

v²=171.67

v=√171.67

v=13.1m/s

A physicist is creating a computational model of a falling person before and after opening a parachute. What boundary conditions would be important here?the air resistance encountered as the person falls

the speed at which the person falls

the change in kinetic and potential energy

the location where potential energy is zero

Answers

Answer:

the location where potential energy is zero

Explanation:

Answer:

Air resistance

Explanation:

Air resistance encountered as the person falls

It's a little hard can u help

Answers

I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades

Why Coulomb force is called "Mutual Force"???????????????????????????????????????

Answers

Answer

The Columb's law is the same as Gravitational law

Explanation

As we see the formula of both Coulomb and Gravitational Law,

F_g = GM_1M_2/R_2     (1)

F_c = kq_1q_2/r_2         (2)

The masses (M) in formula (1) experiencing the force of gravitational pull with each other which varies with changing the distance. In the formula (2), the charges also are felling the forces on each other which varies with distance. The charges and masses are just like the objects which are experiencing the forces which have a common factor as distance. The gravitational force is also called the mutual forces.

А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?​

Answers

Answer:

I am not able to answer this question please don't mind...

Explanation:

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Final answer:

The maximum expected measurement error for a pressure gauge measuring 0-10 bar with an inaccuracy of 1% of full-scale reading is 0.1 bar. When the gauge measures 1 bar, the expected inaccuracy is 10%.

Explanation:

The inaccuracy mentioned here is related to the full-scale reading which means the error is calculated based on the top measurement value. The pressure gauge range is 0-10 bar, so the inaccuracy is one percent of this. (a) Thus, the maximum measurement error expected for this instrument is 1.0% of 10 bar i.e., 0.1 bar. (b) If the gauge is measuring a pressure of 1 bar, then the relative error expressed as a percentage would be the absolute error (0.1 bar) divided by the observed reading (1 bar) i.e., 10%. It means, when measuring 1 bar pressure, the expected measurement error is 10%. This is an example of how instrument inaccuracy is properly interpreted and employed when working with various measurements.

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If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don't have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart

Answers

Complete Question

If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don’t have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mm of Hg (= 0.16 atm = 16 kP = 1.6 × 104 N/m2).

Answer:

The pressure at the brain is P_b  = 89.872 \ mm \ of \ Hg

Explanation:

Generally is mathematically denoted as

                  P = \rho gh

Substituting 1025 kg/m^3 for \rho(the  density) , 9.8 m/s^2 for g (acceleration due to gravity) , 0.4m for h (the height )

We have that the pressure difference between the heart and the brain is

              P = 1025 * 9.8 *0.4

                  = 4018 N/m^2

But the pressure of blood at the heart is given as

               P_h=120 mm of Hg =120 * 133 =  1.59*10^3Pa

Now the pressure at the brain is mathematically evaluated as

                 P_b = P_h - P

                     = 1.596*10^4 - 4018

                     = 11982 N/m^2

                      P_b= (11982)/(133) = 89.872 \ mm \ of \ Hg

   

     

Final answer:

When you stand up quickly, the blood pressure at your brain is lower than at your heart. The decrease in blood pressure can be calculated using the equation ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart.

Explanation:

When you stand up quickly, your blood pressure drops because the blood vessels don't have enough time to expand and compensate for the change in posture. The brain, which is 0.4 m higher than the heart when standing, experiences a decrease in blood pressure. To calculate how much lower the blood pressure is at the brain compared to the heart, we need to use the equation: ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, we can use the height difference of 0.4 m and the density of blood to find the change in pressure.

Using the equation, ΔP = ρgh, we can calculate the change in pressure:

  1. ρ = density of blood = 1060 kg/m³ (approximately)
  2. g = acceleration due to gravity = 9.8 m/s² (approximately)
  3. h = height difference = 0.4 m

Plugging in the values into the equation, we get:

ΔP = (1060 kg/m³)(9.8 m/s²)(0.4 m) = 416.32 Pa

Therefore, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart when standing up quickly.

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